Question

$\int \frac{(x+a{)}^3}{x^3}+\frac{1-x^4}{1-x}-\frac{7}{x\sqrt{(x^2-1)}}+\frac{x+2}{(x^2+1{)}^2}+\frac{(1+x{)}^3}{\sqrt{\left(x\right)}}dx$
=$x+kalogx-k\frac{a^2}{x}-\frac{a^m}{n{x}^2}+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}$

$-7{\sec }^{-1}x+log(x+1)-\frac{1}{x+1}+2\sqrt{x}+2x^{3/2}+\frac{p}{5}{x}^{5/2}+\frac{2}{r}{x}^{7/2}$. Find the value of $k+m+n+p-r$.

Answer 1

Let $I=\int (\frac{{(x+a)}^3}{x^3}+\frac{1-x^4}{1-x}-\frac{7}{x\sqrt{x^2-1}}+\frac{x+2}{{(x^2+1)}^2}+\frac{{(1+x)}^3}{\sqrt{x}})dx$
$=I_1+I_2+I_3+I_4+I_5$
such that
$I_1=\int \frac{x^3+a^3+3x^{2}a+3xa}{x}dx=\int (x^2+\frac{a^3}{x}+3xa+3a)dx=\frac{x^3}{3}+a^3\log x+\frac{3}{2}{x}^{2}a+3ax$
$I_2=\int \frac{(1-x^4)}{1-x}dx$
$=\int \frac{(1-x)(1+x)(1+x^2)}{(1-x)}dx=\int (1+x)(1+x^2)dx$
$=\int (x^3+x^2+x+1)dx=\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}+x$
$I_3=\int \frac{7}{x\sqrt{x^2-1}}dx$
put $\sqrt{x^2-1}=0\Rightarrow \frac{x}{\sqrt{x^2-1}}dx=du$
$\therefore I_3=7\int \frac{1}{u^2+1}du=7{\tan }^{-1}u=7{\tan }^{-1}\left(\sqrt{x^2-1}\right)=-7{\sec }^{-1}x$
$I_4=\int \frac{x+2}{{(x^2+1)}^2}dx$
Put $x=\tan t\Rightarrow dx={\sec }^{2}tdt$
$\therefore I_4=\int {\cos }^{2}t(\tan t+2)dt$
$=\int (-2{\sin }^{2}t+\tan t+{\sin }^{2}t(-\tan t)+2)dt=\frac{(2x-1)}{2(x^2+1)}+{\tan }^{-1}x$
$I_5=\int \frac{{(1+x)}^3}{\sqrt{x}}dx=\int \frac{1+x^3+3x+3x^2}{\sqrt{x}}dx$
$=\int (\frac{1}{\sqrt{x}}+x^{\frac{5}{2}}+3x^{\frac{1}{2}}+x^{\frac{3}{2}})dx=2x^{\frac{1}{2}}+\frac{2x^{\frac{7}{2}}}{7}+\frac{x^{\frac{3}{2}}}{2}+\frac{2x^{\frac{5}{2}}}{5}$
$\therefore k+m+n+p+r=7$